Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.A.
KBH
B.
KB2H4
C.
KB3H9
D.
K2B5H9

Answers

Answer 1

Answer: Use the percent as mass in g
Mass g) --> g/mol
K, B and H molar mass from periodic table

54.93 g K x (1 mol K/ 39.10 g K) = 1.405 mol K
38.73 g B x (1 mol B/ 10.81 g B) = 3.583 mol B
6.34 g H x (1 mol H/ 1.008 g H) = 6.290 mol H

Divide all three answers by the smallest value and you will get
1.000 mol K
2.550 mol B
4.477 mol H
now multiply these three answers by a number that will make all a whole number or (a number with a 9 as the first decimal point)
so multiply by 2
2 mol k
5 mol B
and 9 mol H
E.F. = K2B5H9
Answer is D
Sorry for bad explanation!!!!!!

Answer 2

Answer:

Hello!

Find the empirical formula of a compound which contains 54.93% potassium, 38.73% boron and 6.34% hydrogen.

A. KBH

B. KB2H4

C. KB3H9

D. K2B5H9

We have the following data:

Potassium (K) ≈ 39 a.m.u (g/mol)

Boron (B) ≈ 11 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

K: 54.93 % = 54.93 g

B: 38.73 % = 38.73 g

H: 6.34 % = 6.34 g

The values (in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

$K: (54.93\:\diagup\!\!\!\!\!g)/(39\:\diagup\!\!\!\!\!g/mol) \approx 1.408\:mol$

$B: (38.73\:\diagup\!\!\!\!\!g)/(11\:\diagup\!\!\!\!\!g/mol) \approx 3.521\:mol$

$H: (6.34\:\diagup\!\!\!\!\!g)/(1\:\diagup\!\!\!\!\!g/mol) = 6.34\:mol$

We realize that the values found above are not integers, so we divide these values by the smallest of them, so that the proportion does not change, let us see:

$K: (1.408)/(1.408)\to\:\:\boxed{K = 1}$

$B: (3.521)/(1.408)\to\:\:\boxed{B \approx 2.5}$

$H: (6.34)/(1.408)\to\:\:\boxed{H \approx 4.5}$

convert number of atomic radio into whole number

2 * (1 : 2.5 : 4.5)

= 2 : 5 : 9 ← whole number of atomic radio

K = 2

B = 5

H = 9

Thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{K_2B_5H_9}}\Longleftarrow(Empirical\:Formula)\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:

D. K2B5H9

________________________

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Was this correct? I am pretty sure it is since you are going from a solid to a gas in the first reaction.

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(2) 2 Mg(s) + O2(g) --> 2 MgO(s) solid & gas to a solid

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(1) CaCO3(s) --> CaO(s) + CO2(g) --- solid to solid & gas - most entropy here

(2) 2 Mg(s) + O2(g) --> 2 MgO(s) solid & gas to a solid

(3) H2O(g) --> H2O(l) gas to a liquid

(4) CO2(g) --> CO2(s) gas to a solid

Thus the correct option is 1.

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Answers

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Answers

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